# Proving Trig Identities

• Apr 23rd 2009, 02:19 PM
casey_k
Proving Trig Identities
I'm having trouble making sense of addition and subtraction in trig identities.

cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x
• Apr 23rd 2009, 02:49 PM
Prove It
Quote:

Originally Posted by casey_k
I'm having trouble making sense of addition and subtraction in trig identities.

cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x

Notice that $\displaystyle \cos^6{x} = (\cos^2{x})^3$

$\displaystyle = (1 - \sin^2{x})^3$

$\displaystyle = \sum_{k = 0}^{3}\left(_k^n\right)1^{n - k}+(-\sin^2{x})^k$

$\displaystyle = 1 - 3\sin^2{x} + 3\sin^4{x} - \sin^6{x}$.

So we can rewrite the original function as

$\displaystyle 1 - 3\sin^2{x} + 3\sin^4{x} - \sin^6{x} + \sin^6{x}$.

What does this equal?
• Apr 23rd 2009, 02:51 PM
skeeter
Quote:

Originally Posted by casey_k
I'm having trouble making sense of addition and subtraction in trig identities.

cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x

two hints ...

$\displaystyle (a^2)^3 + (b^2)^3 = (a^2 + b^2)(a^4 - a^2b^2 + b^4)$

$\displaystyle \cos^2{x} = 1 - \sin^2{x}$
• Apr 23rd 2009, 04:13 PM
casey_k
I didn't get it at first because I expanded the right side both (cos^2x)^3 + (sin^2x)^3 but then I realized I only had to expand cos.

Thank you!